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How do you find the limit of #x/sqrt(x^2-9)# as #x->3^-#?

Answer 1

The limit does not exist.

If we are staying in the real numbers, then the domain of the function is #(-oo,-3) uu (3,oo)# so we cannot approach positive 3 from the left.

If we are allowed to use imaginary numbers, then the limit still does not exist, because as x approaches 3 from the left, the numerator goes to 3 and the denominator goes to #0# (through purely imaginary values).

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Answer 2

Luke Alvarez

To find the limit of x/sqrt(x^2-9) as x approaches 3 from the negative side, we can substitute the value of x into the expression. However, this would result in an undefined value since the denominator becomes zero.

To evaluate this limit, we can simplify the expression by multiplying both the numerator and denominator by the conjugate of the denominator, which is sqrt(x^2-9) + 3.

After simplifying, we get (x * (sqrt(x^2-9) + 3)) / (x^2 - 9).

Now, as x approaches 3 from the negative side, we substitute this value into the simplified expression.

The limit of x/sqrt(x^2-9) as x approaches 3 from the negative side is 3/0, which is undefined.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #x/sqrt(x^2-9)# as #x-&gt;3^-#?

How do you find the limit of #x/sqrt(x^2-9)# as #x->3^-#?